Work Required

1. Feb 8, 2009

shimizua

1. The problem statement, all variables and given/known data
Calculate the work required to move a charge of -0.51x10^-12 C from i' to b'.

2. Relevant equations
i know that the equation is -W=qEl and that it can also be expressed as -W=-q$$\Delta$$V or W=q$$\Delta$$V

3. The attempt at a solution
well q is given as -0.51x10^-12
the problem i am having is how do i find $$\Delta$$V without knowing E? cause i know i can figure out l just by using the picture. So if you can just help me with find out how to get $$\Delta$$V that would be great!

2. Feb 8, 2009

LowlyPion

Looks like your ΔV is going from +3 to -3. That looks like a Voltage drop of 6v from the image, or ΔV = -6.

Just count the contours is my method.

3. Feb 8, 2009

Redbelly98

Staff Emeritus
The figure is a contour plot of the potential. Each contour line represents a constant potential value.

The figure also indicates which contour lines correspond to -5V, 0V, and +5V. Using that, can you say:
What is the voltage at point i?
What is the voltage at point b?

4. Feb 9, 2009

shimizua

ok so i did the (-.51x10^-12)(-6) and got an answer of 3.06x10^-12 J and it was said to be wrong. Where did i got wrong?

5. Feb 9, 2009

shimizua

oh and i was given this hint when i put my answer in

The equipotential lines shown are separated by 1 kV. (NOTE! That's kV, not volts!) Work to move a charge is the increase in potential energy of the charge. The potential energy is the potential times the charge.

6. Feb 9, 2009

cjl

The answer remains the same though, whether it is kV or V. It's still a matter of counting contour lines.

7. Feb 9, 2009

Redbelly98

Staff Emeritus
If it's actually 1 kV, and not 1 V, for the contour intervals, then instead of 6V it is really ____?

8. Feb 9, 2009

shimizua

ok so then it would mean that it was really 6 kV or 6000 V and then when i did that i got 3x10^-9 J and that also came up wrong

9. Feb 9, 2009

Newton V

10. Feb 9, 2009

shimizua

haha, it is the graph that they gave up. i didnt make it

11. Feb 9, 2009

Newton V

ok i give up... i am having my own probelms...

12. Feb 9, 2009

LowlyPion

It's plus Work moving an opposite charge away from + and toward a - V.

If the contours are 103v then it's 3.06 * 10-9 J or 306 ergs.

If it's not precision or units then I've no idea what is asked.

13. Feb 10, 2009

shimizua

i got the answer. i did the 3.06e-09 J and it seemed to work even though i did 3.1e-09 before. i guess it had to be 06. thanks for all your help guys